Elementary 3 chemistry questions about alkalis, 3 chemical alkalis

1.Chemical formula for neutralization reaction of base and acid:

1 Hydrochloric acid and caustic soda react: HCl + NaOH ==== NaCl + H2O

2.Hydrochloric acid and potassium hydroxide reaction: HCl + KOH ==== KCl + H2O

3 Hydrochloric acid and copper hydroxide reaction: 2HCl + Cu(OH)2 ==== CuCl2 + 2H2O

4.Hydrochloric acid and calcium hydroxide reaction: 2HCl + Ca(OH)2 ==== CaCl2 + 2H2O

5.Hydrochloric acid and iron hydroxide reaction: 3HCl + Fe(OH)3 ==== FeCl3 + 3H2O

6.Aluminum hydroxide drug ** Hyperacidity: 3HCl + AL(OH)3 ==== ALCL3 + 3H2O

7.Sulfuric acid and caustic soda reaction: H2SO4 + 2NaOH ==== Na2SO4 + 2H2O

8.Sulfuric acid and potassium hydroxide reaction: H2SO4 + 2KOH ==== K2SO4 + 2H2O

9.Sulfuric acid and copper hydroxide reaction: H2SO4 + Cu(OH)2 ==== CuSO4 + 2H2O

10.Sulfuric acid and iron hydroxide reaction: 3H2SO4 + 2Fe(OH)3==== Fe2(SO4)3 + 6H2O

11.Nitric acid and caustic soda reaction: HNO3 + NaOH ==== Na3 + H2O

2.Chemical formula by which a base reacts with a non-metallic oxide:

1 Caustic soda deteriorates when exposed to air: 2NaOH + CO2 ==== Na2CO3 + H2O

2 Caustic soda absorbs sulfur dioxide gas: 2NaOH + SO2 ==== Na2SO3 + H2O

3 Caustic soda reacts with sulfur trioxide: 2NaOH + SO3 ==== Na2SO4 + H2O

4 Slaked lime deteriorates when placed in air: Ca(OH)2 + CO2 ==== CaCO3 + H2O

5.Slaked lime absorbs sulfur dioxide: Ca(OH)2 + SO2 ==== CaSO3 + H2O

3.Chemical formula by which a base reacts with soluble salts:

1.Sodium hydroxide vs copper sulfate: 2NaOH + CuSO4 = Cu(OH)2 + Na2SO4

2.Sodium hydroxide with ferric chloride: 3NaOH + FeCl3 = Fe(OH)3 + 3NaCl

3 Sodium hydroxide with magnesium chloride: 2NaOH + MgCl2 = Mg(OH)2 + 2NaCl

4.Sodium hydroxide with copper chloride: 2NaOH + CuCl2 = Cu(OH)2 + 2NaCl

5.Calcium hydroxide vs sodium carbonate: Ca(OH)2 + Na2CO3 = CaCO3 + 2NaOH

It's too simple Ca(OH)2+H2SO4=CASO4+2H2OSO2+CA(OH)2=CASO3+H2O (calcium-based sulfur fixation method used in sulfuric acid plant to treat waste gas, not in the book but on the paper when doing the question, and your teacher also talked about it, sulfur dioxide and alkali form sulfite).

3NaOH + FeCl3 = Fe(OH)3 + 3NaCl remember. The conditions under which the metathesis reaction takes place are to generate a precipitate or water or gas, and the valency of the ions exchanged with each other is unchanged.

The alkali in junior high school mainly talks about sodium hydroxide, calcium hydroxide, and acid mainly talks about dilute sulfuric acid and dilute hydrochloric acid. You take these kinds of things, plus non-metallic oxides, mainly carbon dioxide sulfur dioxide and sulfur trioxide, and slowly synthesize them themselves, and if there is no water or gas or precipitation generation, the reaction cannot take place. The solubility table is in Appendix II of the book

+H+ = H2O (ionic) Base + Acid = Salt + Water (General Formula) NaOH + HCl = NaCl + H2O (specific example).

2.Bases do not have the universality to react with non-metal oxides.

Metal cation = neoalkali (ionic) base + salt = neoalkali + neosalt (general formula) 2NaOH + CuCl2 = Cu(OH)2 + 2NaCl Ca(OH)2 + Na2CO3 = 2NaOH + CaCO3 (specific example) [The base and salt of the reactant must be soluble, and the base or salt of the product must have a poorly soluble].

naoh+hcl=nacl+h2o

ca（oh）2+h2so4=caso4+2h2ohcl+koh=kcl+h2o

Ca(OH)2+CO2=CaCO3 (white precipitate) + H2O2NaOH+CO2=Na2CO3+H2O

2koh+co2=k2co3+h20

Na2CO3 + Ca(OH)2 = CaCO3 (white precipitate) + 2NaOHoKhkoH + KhCO3 = K2CO3 + H2O

2NaOH + CuSO4 = Cu(OH)2 (blue precipitate) + Na2SO4NaOH + NaHCO3 = Na2CO3 + H2O

1. hcl+koh=kcl+h2o

2. 2koh+co2=k2co3+h203.koh+khco3=k2co3+h2oI'm sorry, I can't type the number, I barely understand it.

The most common bases in junior chemistry are:

1.Sodium hydroxide (NaOH).

2.Calcium hydroxide (Ca(OH)2).

Other alkalis include ammonia and aluminum hydroxide.

Potassium hydroxide, iron hydroxide, copper hydroxide, barium hydroxide, magnesium hydroxide, iron hydroxide, ferrous hydroxide are all involved.

a Hydrochloric acid.

First of all, it can be seen that the purple color is litmus.

Add litmus drops to the remaining four solutions, the one that turns red is hydrochloric acid, and the remaining three will turn blue.

The identified droplets of hydrochloric acid were added to the remaining three solutions, and the bubbles appeared in sodium carbonate.

The identified sodium carbonate droplets were added to the remaining two solutions, and a white precipitate appeared calcium hydroxide.

The last one left is sodium hydroxide.

Therefore, the order of identification is: litmus, hydrochloric acid, sodium carbonate, calcium hydroxide, sodium hydroxide.

Are there any other reagents? If you have it, it's simple:

Hydrochloric acid: It is the only acidic substance among the three substances, so litmus reagent is used, and if the reagent turns red, it is hydrochloric acid.

Then distinguish the clarified lime water (Ca(OH)2): carbon dioxide (CO2): CO2 + Ca(OH)2 CaCO3 + H2O

If the reagent is turbid, it is clarified lime water.

Sodium hydroxide: add magnesium chloride (MgCl2): MgCl2 + 2NaOH 2NaCl + Mg(OH)2 to the reagent

If there is a white spot (magnesium hydroxide), it is sodium hydroxide. There should be a reagent that contains Na (sodium), so it can.

With the flame color reaction, some reagents are dotted on the platinum wire and placed on the flame, if the flame is yellow, it is sodium hydroxide.

Add sodium carbonate.

There is gas in hydrochloric acid.

There was no change in sodium hydroxide.

There is precipitation in the clarified lime water.

Sufficient amount of CO2 is introduced, and there is no obvious change in hydrochloric acid, sodium hydroxide is precipitated first and then dissolved, and clarified lime water is always precipitated.

Answer: c

If the amount of iron powder is too much, there may be iron in the insoluble matter, so a is not true;

If the amount of iron is just enough to react with sulfuric acid and copper sulfate, then there is no iron powder left, and the insoluble matter is copper, so b is not right;

If there is copper sulfate in the solute, then there will definitely be copper precipitation after the iron is put in, so d is not right.

Choosing C, putting in the iron nail, there is no change, indicating that there is no CuSO4 in the solution, then Cu

In ** went to it It is obvious that there is an excess of iron in front of it Cu is replaced, so there must be Cu in the intolerable matter To what extent is the excess of iron here is unknown, it is possible that iron is still left after the reaction with dilute sulfuric acid and Cuso4 So it is unknown.

Because there is no change in the iron set added later, there must be no copper in the filtrate, so D is wrong. At the same time, it can be proved that the iron powder must replace all the copper, which may be sufficient or excessive. So there must be Cu in the insoluble matter, and there may be iron.

There must be Cu and iron in the insoluble matter of C.

50ml*

60g*40%=(60g+x)*10%

x=180g

1) To prepare hydrochloric acid with a solute mass fraction of 10% with 50ml of concentrated hydrochloric acid, 180g of water needs to be added

1.The chemical equation for the reaction of metal R is R+ 2HCl === RCl2 + H2

r+ 2hcl === rcl2 + h2↑x 73 x+71 2n 73g*10% m

73/x=56

73/m=73/n=

2.The mass of the hydrogen gas generated.

3.The elemental symbol for the metal R is (Fe) [a divalent metal with a relative atomic mass of 56, possibly iron].

4.The purity of the metal.

1) Because the density of 10% hydrochloric acid is not known, it is difficult to solve. However, it is generally accepted that the density of dilute hydrochloric acid is.

It should also be assumed that the mixed volume of hydrochloric acid and water remains the same.

Solve the problem under the above assumptions:

If the volume of hydrochloric acid prepared is xml, then there is: 50ml*, and the solution is x=240.

Therefore, the volume of water that needs to be added is: 240ml-50ml=190ml.

2）1。Because - is r is bivalent, so the equation for the reaction of r with hydrochloric acid is: r + 2HC - r Cl2 + H2. (Hydrogen with an upward arrow).

2。Because the hydrochloric acid is completely reactive, and the amount of hydrochloric acid is: (73g*10%) (. Therefore, the hydrogen produced is, ie.

3。Because the impurities are insoluble in water and acid, the filtered evaporated solid powder is Rcl2, the mass is, and the amount of the substance is obtained according to the complete reaction of hydrochloric acid), so the molar mass of R is:. Therefore, the metal is iron (Fe).

4。From the complete reaction of hydrochloric acid and iron, the mass of iron is :, so the purity of iron is:

Seeing that MCL3 first reacted to ALCL3, and then calculated, sure enough! Looking at the generation of hydrogen 2g, the reaction ratio of Al to H2 is 2:3, so it should be two-thirds Al, 27*(2 3)=18g, which is calculated to be 18g

A few drops of phenolphthalein test solution are dropped into the beaker containing NaOH solution, the test solution becomes (pink) color, at this time the solution (pH is greater than 7), dilute hydrochloric acid is added to this solution drop by drop, and continue to **, until the test solution is just colorless, the solution is (medium), pH (- is 7), continue to drop dilute hydrochloric acid, the solution is (no) color, pH (less than 7)-, in the whole experimental process, the solution undergoes a (neutralization) reaction, the chemical equation is HCl+NaOH=NaCl+H2O

Red is greater than 7

Medium is equal to 7 and none is less than 7

Neutralize NaOH + HCl = NaCl + H2O

Red pH>7 neutral pH=7 Colorless pH<7 neutralizing reaction NaOH+HCl=NaCl+H2O

Pink, 7, Medium, =7, Colorless, 7, Acid-base neutralized, NaOH+HCl=NaCl+H2O

Red, pH 7, this can't be calculated without data.

Neutral, 7 colorless, 7

naoh+hcl=nacl+h2o